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+\noindent{\bf 6.4\#2:}
+  $$\vec x' = \pmatrix{2&-4&-1\cr
+                       1&1&3\cr
+                       3&-4&-2\cr} \vec x$$
+  
+  Characteristic Polynomial: $-\lambda^3 + \lambda^2 - 15\lambda - 17 = -(x+1)(x-(1+4i))(x-(1-4i))$
+
+  Eigenvalues and eigenvectors: $\pmatrix{-1\cr-1\cr1\cr}$ for $-1$, $\pmatrix{1\cr-i\cr1\cr}$ for $1+4i$, $\pmatrix{1\cr i\cr1\cr}$ for $1-4i$.
+
+$e^te^{4it}\vec v_2 = e^t(\cos 4t + i\sin 4t)\pmatrix{1\cr-i\cr1\cr} \to \vec x = c_1\pmatrix{-1\cr-1\cr1\cr}e^{-t} + c_2\pmatrix{\cos 4t\cr\sin 4t\cr\cos 4t}e^t + c_3\pmatrix{\sin 4t\cr-\cos 4t\cr\sin4t\cr}e^t$
+
+\noindent{\bf 4.1\#16:}
+
+%%A spring is stretched 10 cm by a force of 3 newtons (N). A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 N when the velocity of the mass is 5 m/s. The mass is pulled down 5 cm below its equilibrium position and given an initial upward velocity of 10 cm/s. 
+
+Starting from the general formula $my''(t) + \lambda y'(t) + ky(t)=0$, $k = {3\over.1}$, $\lambda = {3\over.05}$, $m = 2$. It is assumed that gravity is downwards, so $y'(0) < 0$ because it starts with upward velocity.
+
+$2y''(t) - 60y'(t) - 30y(t) = 0$, $y(0) = 5$ (downwards position), $y'(0) = -10$. %%Signs might be wrong
+
+\noindent{\bf 4.2\#28:}
+
+$ay'' + by' + cy = 0$ where $b^2-4ac=0$, $y_1(t) = e^{-bt/2a}$.
+
+Eigenvalues are the same by the quadratic formula, so because $y' = y_1$, $te^{-bt/2a} = y_2$.
+
+\noindent{\bf 4.3\#48:}
+
+$y'' + (3-\alpha)y' - 2(\alpha-1)y = 0$.
+
+Eigenvalues: ${\alpha-3\pm\sqrt{\alpha^2+2\alpha+1}\over 2} = {\alpha-3\pm(a+1)\over2}$
+
+Where $\alpha-3-(\alpha+1) = -4 > 0$, solutions trend to $\infty$, so none do.
+
+Where $\alpha-3+(\alpha+1) = 2\alpha-2 > 0 \to \alpha > 1$, solutions trend to $0$.
+
+\bye